Silverstone quotes 166 tons in Warships of the Civil War Navies for the Hamptons, taken from ORNs, but without defining the figure. They are shown as 106ft x 21ft x 5ft, but again with no definition as what these measurement are. Tonnage measurements at this time were a nightmare anyway as there was no such thing as standard measurement. I reckon by rule of thumb the 166 tons is about right.

Silverstone quotes 166 tons in Warships of the Civil War Navies for the Hamptons, taken from ORNs, but without defining the figure. They are shown as 106ft x 21ft x 5ft, but again with no definition as what these measurement are. Tonnage measurements at this time were a nightmare anyway as there was no such thing as standard measurement. I reckon by rule of thumb the 166 tons is about right.

Thank you John. I'm engaged in some jackleg reverse engineering and am collecting data to create a spread on block coefficients for selected small hulls including the Crimean gunboats, and several of the Porter variants in the same range as the Maury boats. The goal is to get an educated guess on the displacements of the Singer torpedo boats. A major clinker is without a three view you don't know how they handled the submerged portion of the hull. If they were influenced by the Porter designs I would expect something similar to a flat bottom. Just for the heck of it I calculated the block volume for a rectangle with the length, beam and waterline of the Texas boats, then made a guesstimate of a block coefficient of .75 and ended up with about 216 tons. This may be an over estimate and without more data it seems smarter to create a parametric table based on block coefficients for hulls of known vessels with similar dimensions. The problem is that most of them had considerably more beam versus length. Some of the blockade runners might be closer geometrically, but I think the build conditions didn't lend themselves to elegant hulls. What displacement do you get with your rule of thumb for a hull 114 ft long, 14 ft beam and 7 ft 9 inches depth of hold with a draft of 5 ft 9 inches?

Thank you John. I'm engaged in some jackleg reverse engineering and am collecting data to create a spread on block coefficients for selected small hulls including the Crimean gunboats, and several of the Porter variants in the same range as the Maury boats. The goal is to get an educated guess on the displacements of the Singer torpedo boats. A major clinker is without a three view you don't know how they handled the submerged portion of the hull. If they were influenced by the Porter designs I would expect something similar to a flat bottom. Just for the heck of it I calculated the block volume for a rectangle with the length, beam and waterline of the Texas boats, then made a guesstimate of a block coefficient of .75 and ended up with about 216 tons. This may be an over estimate and without more data it seems smarter to create a parametric table based on block coefficients for hulls of known vessels with similar dimensions. The problem is that most of them had considerably more beam versus length. Some of the blockade runners might be closer geometrically, but I think the build conditions didn't lend themselves to elegant hulls. What displacement do you get with your rule of thumb for a hull 114 ft long, 14 ft beam and 7 ft 9 inches depth of hold with a draft of 5 ft 9 inches?

Assuming that's OA x EX x MAX, 131.1 tons imperial, 150 tonnes metric. Incidentally using my formula gives 159 tons for the Maury /porter boats.
Most builders were working to the wave line theory based on centuries of experience. The first ironclad I can find that didn't oddly is Tift's Mississippi, whose entry and exit lines are closer to a modern definition, except that the run aft is usually lifted vertically over varying distances depending on the vessels intended purpose and speed.

Assuming that's OA x EX x MAX, 131.1 tons imperial, 150 tonnes metric. Incidentally using my formula gives 159 tons for the Maury /porter boats.
Most builders were working to the wave line theory based on centuries of experience. The first ironclad I can find that didn't oddly is Tift's Mississippi, whose entry and exit lines are closer to a modern definition, except that the run aft is usually lifted vertically over varying distances depending on the vessels intended purpose and speed.

Length overall is from the furthest point of the hull forward to the furthest point aft.
Extreme beam is the distance over the hull width between outside planking /plating on each side. Must not includes sponsons.
Maximum Draught is the maximum depth between waterline and keel.
l

Assuming that's OA x EX x MAX, 131.1 tons imperial, 150 tonnes metric. Incidentally using my formula gives 159 tons for the Maury /porter boats.
Most builders were working to the wave line theory based on centuries of experience. The first ironclad I can find that didn't oddly is Tift's Mississippi, whose entry and exit lines are closer to a modern definition, except that the run aft is usually lifted vertically over varying distances depending on the vessels intended purpose and speed.

Hi John. by any chance are the dimensions you are putting in metric? and are you dividing the result by the number of pounds in a ton?
I checked my conversion tables and they claim that an Imperial ton is 2240 lbs or 1.12 short (US) tons. So I took your 131.1 tons imperial and multiplied this by 1.12 and get 146.8 shot tons. Does this compute?

George, no, they are our feet and inches ! I used 2,000 lbs as the short ton, obviously your US short ton is slightly different. How many lbs is a US short ton?
Mentioning feet and inches, you have to beware of measurements in European documents in original languages before the metric system, for example the Swedish foot and inches were bigger than ours.

Sorry left out my formula, The idea, faced with so many conflicting weight measurements was to find a way of getting a standard comparison of size. This was before the computer and the electric calculator, so I spent many hours with a hand cranked Burroughs (how many people remember those) and many many sheets or paper with as many dimensions and tonnages as I could find.

It is : Length overall x extreme beam x maximum draught in feet (inches rounded up or down as appropriate if you wish, although a calculator negates the need to do that)
Divide the result by 70.
So Porter's 150ft gunboat plan gives:
172ft OA x 45ft EX x 12ft MAX
-------------------------------------- = 1,327 imperial tons displacement.
70